0
$\begingroup$

For Spin 1/2 particles, the spin operator along an arbitrary axis defined by the normalized vector $\vec{n}$ is given by a weighted sum over the pauli matrices:

$$S(\vec{n})=n_x \sigma_x + n_y \sigma_y + n_z \sigma_z.\tag{I}$$

If a particle has been prepared in an up-eigenstate $\chi_n$ with respect to $S(\vec{n})$ and then we measure the spin again along a different axis $\vec{m}$, the probability that we will measure spin up again is given by the simple formula

$$|\langle \chi_m|\chi_n\rangle|^2=\cos\left(\frac{\theta}{2}\right)^2.\tag{II}$$

where $\theta$ is the angle between the two axes.

This formula can be derived by considering the general up-state eigenvector of $S(\vec{n})$, then calculating it`s inner product with another general eigenvector of $S(\vec{m})$ and then taking the absolute square, which is quite a mess of a calculation.

My question is: Is there an easy/elegant way to derive Eq. II that rests on as few assumptions as possible?

$\endgroup$
0
$\begingroup$

Since $|\langle \chi_m|\chi_n\rangle|^2$ is a scalar, we can without loss of generality take $\vec n$ to be along $+\hat z$ and $\vec m$ to be $R_z(\varphi)R_y(\theta)\vec n$. Then $\vert \vec n\rangle=\vert +\rangle$ and $\vert \vec m\rangle = R_z(\varphi)R_y(\theta) \vert +\rangle$. From this it immediately follows that $$ \langle + \vert \vec m\rangle = \langle + \vert R_z(\varphi)R_y(\theta) \vert +\rangle = e^{-i\varphi/2}\cos\left(\frac{\theta}{2}\right) $$ and your result follows immediately.

For a general orientation, you can replace $\vert +\rangle$ by $R(\omega)\vert +\rangle$ and $\vert \vec m\rangle $ by $R(\omega)\vert m\rangle$ with $R(\omega)$ an arbitrary rotation as this does not affect the relative angle between the vectors. Then $$ \langle + \vert R^{-1}(\omega)R(\omega)\vert \vec m\rangle = \langle +\vert \vec m\rangle\, , $$ and the general result follows.

Note that, for spin-1/2, $R_z(\varphi)=e^{-i\varphi\sigma_z/2}$ so $\langle +\vert R_z(\varphi)= e^{-i\varphi/2}\langle +\vert$ and that likewise $$ R_y(\theta)\vert +\rangle=e^{-i\theta \sigma_y/2}\vert +\rangle =\cos\left(\frac{\theta}{2}\right) \vert +\rangle + \sin\left(\frac{\theta}{2}\right) \vert -\rangle\, , $$ using elementary properties of $\sigma_y$.

For vectors we have $R_z(\varphi)=e^{-i\varphi L_z}$ and $R_y(\theta)=e^{-i\theta L_y}$ with $L_z$ and $L_y$ the $3\times 3$ matrices representing the appropriate generators.

$\endgroup$
  • $\begingroup$ So i guess the only thing we needed to assume here is how the rotation operator that transforms a 3D vector acts upon the corresponding ket vector?Is there a simple way to see that it has to be a multiplication by exp(-i phi/2)*C0s(th/2) ? Also, is it true that in the way you defined it, |+> = [1,0] (the eigenvector of sigma z)? $\endgroup$ – curio Aug 18 '17 at 19:39
  • $\begingroup$ @curiosity I added some additional details. $\endgroup$ – ZeroTheHero Aug 18 '17 at 19:43
  • $\begingroup$ @curiosity ... and yes $\vert +\rangle$ is [1,0] $\endgroup$ – ZeroTheHero Aug 18 '17 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.